704 - Problems 7‐1 through 7‐3 were deleted since you...

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Unformatted text preview: Problems 7‐1 through 7‐3 were deleted, since you will not be held responsible for the nomenclature of amines on the final exam. Practice Problem 7‐4 – Answer Propose a structure for Compound A (C5H13N), given its IR and 1H NMR spectra (not reproduced in this answer key). Practice Problem 7‐5 – Answer Which of the two nitrogens in pyridoxamine (a form of vitamin B6) is the stronger base? Explain your reasoning. The nitrogen atom highlighted in the box to the left is the stronger base. This nitrogen atom is sp3 hybridized, while the other N atom is sp2 hybridized. The sp2 hybridized N atom has a greater percentage s character, so the lone pair electrons are held closer to the nucleus and are therefore less available for interaction with a proton. Practice Problem 7‐6 – Answer Select the stronger base from each pair of compounds. Practice Problem 7‐7 – Answer Epibatidine, a colorless oil isolated from the skin of the Ecuadorian poison frog Epipedobates tricolor, has several times the analgesic potency of morphine. (a) Which of the two nitrogen atoms is the more basic? The N atom highlighted in the box is more basic. See the explanation for Problem 7‐5 for the reason. (b) Mark all chiral centers in this molecule The chiral centers are marked with asterisks. * * * Practice Problem 7‐8 – Answer Suppose that you have a mixture of these three compounds. Devise a chemical procedure based on their relative acidity or basicity to separate and isolate each in pure form. These molecules can be separated by extraction into different aqueous solutions. First, the mixture is dissolved in an organic solvent (such as Et2O) in which all three compounds are soluble. Then, the ether solution is extracted with dilute aqueous HCl. Under these conditions, 4‐methylaniline, a weak base, is converted into its protonated form, and will dissolve in the aqueous solution. The aqueous solution is separated, treated with dilute NaOH, and the water‐insoluble 4‐methylaniline precipitates. The ether solution containing the other 2 components is then extracted with dilute aqueous NaOH. Under these conditions, 4‐methylphenol, a weak acid, is converted into its deprotonated form, and will dissolve in the aqueous solution. Acidification of this aqueous solution with dilute HCl forms water‐insoluble p‐cresol, which precipitates. Evaporation of the solvent from the remaining ether solution gives the 4‐nitrotoluene, which is neither acidic nor basic. Practice Problem 7‐9 – Answer Show how to convert each starting material into benzylamine in good yield. Practice Problem 7‐10 – Answer Propose a synthesis of 1‐hexanamine from the following. (a) A bromoalkane of six carbon atoms (b) A bromoalkane of five carbon atoms Practice Problem 7‐11 – Answer How might you prepare pentylamine from the following starting materials? (a) Pentanamide (b) Pentanenitrile (c) 1‐Butene (d) Hexanamide (e) 1‐Butanol (f) 5‐Decene (g) Pentanoic acid Practice Problem 7‐12 – Answer Propose steps for the following conversions using a reaction of a diazonium salt in at least one step of each conversion. (a) Toluene to 4‐methylphenol (p‐cresol) (b) Toluene to p‐cyanobenzoic acid (c) Acetanilide to p‐aminobenzylamine (d) 3‐Methylaniline to 2,4,6‐tribromobenzoic acid (e) Nitrobenzene to 3‐bromophenol (f) Phenol to p‐iodoanisole (g) Toluene to 4‐fluorobenzoic acid Practice Problem 7‐13 – Answer Show how to bring about each step in this synthesis of the herbicide propranil. Practice Problem 7‐14 – Answer Show how to bring about each step in the following synthesis. Practice Problem 7‐15 – Answer Show how to bring about this synthesis. Practice Problem 7‐16 – Answer Show how to bring about each step in the following synthesis. Practice Problem 7‐17 – Answer Methylparaben is used as a preservative in foods, beverages, and cosmetics. Provide a synthesis of this compound from toluene. Practice Problem 7‐18 – Answer Several diamines are building blocks for the synthesis of pharmaceuticals and agrochemicals. Show how both 1,3‐ propanediamine and 1,4‐butanediamine can be prepared from acrylonitrile. Acrylonitrile is a Michael acceptor that can react with ammonia or the cyanide anion. ...
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